Duality Principle (Boolean Algebras)
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This proof is about Duality Principle in the context of Boolean Algebra. For other uses, see Duality Principle.
Theorem
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then any theorem in $\struct {S, \vee, \wedge, \neg}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem.
Proof
Let us take the axioms of a Boolean algebra $\struct {S, \wedge, \vee, \neg}$:
\((\text {BA}_1 0)\) | $:$ | $S$ is closed under $\vee$, $\wedge$ and $\neg$ | |||||||
\((\text {BA}_1 1)\) | $:$ | Both $\vee$ and $\wedge$ are commutative | |||||||
\((\text {BA}_1 2)\) | $:$ | Both $\vee$ and $\wedge$ distribute over the other | |||||||
\((\text {BA}_1 3)\) | $:$ | Both $\vee$ and $\wedge$ have identities $\bot$ and $\top$ respectively | |||||||
\((\text {BA}_1 4)\) | $:$ | $\forall a \in S: a \vee \neg a = \top, a \wedge \neg a = \bot$ |
It can be seen by inspection that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout, does not change the axioms.
Thus, what you get is a Boolean algebra again.
Hence the result.
$\blacksquare$
Sources
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.5$: Theorem $1.14$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Boolean algebra
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): duality: 2.
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Boolean algebra
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): duality: 2.